PSTAT 120A, Spring 2010, Practice Final.
(1) (a) Let
X
and
Y
be independent random variables each having the probability
mass function
p
(
x
) =
1
2
±
2
3
²
x
,
x
= 1
,
2
,
3
,...
Find
P
(
X
= 1
,Y
= 3) and
P
(
X
+
Y
= 3).
(b) Let
X
and
Y
be two independent random variables with the same
probability density function given by
f
(
x
) =
(
e

x
if 0
< x <
∞
0
elsewhere.
Show that
g
, the probability density function of
X/Y
, is given by
g
(
t
) =
(
1
(1+
t
)
2
if 0
< t <
∞
0
t
≤
0
.
Solution:
(a) Since
X
and
Y
are independent with same p.m.f.,
P
(
X
= 1
,Y
= 3) =
P
(
X
= 1)
·
P
(
Y
= 3) =
p
(1)
p
(3) =
1
4
±
2
3
²
4
= 4
/
81
.
P
(
X
+
Y
= 3) =
P
(
X
= 1
,Y
= 2) +
P
(
X
= 2
,Y
= 1) = 2
·
1
4
±
2
3
²
3
= 4
/
27
.
(b) Since
X
and
Y
are independent with the same pdf, the joint density
function for
X
and
Y
is
f
X,Y
(
x,y
) =
f
(
x
)
·
f
(
y
)
i.e.
f
X,Y
(
x,y
) =
(
e

(
x
+
y
)
if 0
< x <
∞
,
0
< y <
∞
0
elsewhere.
We ﬁrst ﬁnd the distribution function for
Z
=
X/Y
. For
z >
0,
F
Z
(
z
) =
P
(
Z
≤
z
)
=
P
(
X/Y
≤
z
) =
ZZ
{
(
x,y
):
x
≤
zy
}
f
X,Y
(
x,y
)
dxdy
=
Z
∞
0
Z
zy
0
e

x
e

y
dxdy
=
Z
∞
0
(
e

y

e

(
z
+1)
y
)
dy
=

e

y
+
e

(
z
+1)
y
z
+ 1
³
³
³
³
∞
0
= 1

1
z
+ 1
.
f
X/Y
(
z
) =
d
dz
F
Z
(
z
) =
1
(
z
+ 1)
2
for
z >
0
and
f
X/Y
(
z
) = 0 for
z
≤
0.
1